# What is the law of multiple proportions?

# Law of multiple proportions

The **law of multiple proportions** states that if two elements form more than one compound when they react with each other, then the proportion of the masses with which one of them combines with a fixed mass of the other, is equal to a ratio of small integers .

The statement of the law can seem complex if you do not have an example at hand. So consider some Oreos, made up of two chocolate tops and a strip of sweetened cream: T _{2} C (T = top and C = cream). If we wanted to invent a more robust Oreo cookie, we would add another strip of cream, in order to have double cream (T _{2} C _{2} or TC).

We could also add another one, so that the cookie has three times more cream than a conventional cookie (T _{2} C _{3} ). What if we put another chocolate cap in the middle of the white stripes (T _{3} C _{2} )? The options are limitless; but we always add a unit of tapa or cream. We do not think of a half lid (1/2 T), or a fifth of cream (1/5 C), as it would be inappropriate.

In the same way it happens with chemical elements: their atoms do not split to form compounds. Therefore, the masses of T or C between their compounds are “always” in a simple relationship.

__Explanation__**Proportions**

The law of multiple proportions, together with the law of definite proportions, preceded stoichiometry and the first chemical formulas. Let’s forget about cookies, but keep their symbols: T and C. Through experiments, elements T and C are found to form various compounds: T _{2} C, TC, and T _{2} C _{3} .

Before chemical formulas it was impossible to know at once what the proportions of the masses of T and C really were in such compounds. They had to be determined first. In one compound, the mass of T was found to double that of C; i.e. 2 grams of T is combined with 1 gram of C.

Then, in the other compound, the masses of T and C were equal: 2 grams of T now combine with 2 grams of C. Here the question arises: what if T and C can still form another compound? If so, it would surely be formed starting with 2 grams of T, but this time they would be combined with 3 grams of C (1 gram + 1 gram + 1 gram).

**Simple mass relations**

The proportions of the masses with which T and C react allow to establish their chemical formulas: T _{2} C (2 grams T: 1 gram C), TC (2 grams T: 2 grams C) and T _{2} C _{3} (2 grams T: 3 grams C). If we want to compare the relations of the masses of T or C in these compounds, it is necessary that one of their masses remains constant; in this case that of T: 2 grams.

Therefore, we will determine the mass proportions of C in these three compounds:

- T
_{2}C: 1 gram C / 2 grams T - TC: 2 grams C / 2 grams T
- T
_{2}C_{3}: 3 grams C / 2 grams T

We will thus have a ratio for the mass of C equal to 1: 2: 3. That is, there are 2 times more C in TC than in T _{2} C, and 3 times more C in T _{2} C _{3} than in T _{2} C. As can be seen, 1: 2: 3 are small integers (they do not even exceed the ten).

__Examples of the law of multiple proportions__For the following examples, the same previous steps will be applied, but we will take into account the molar masses of the respective elements, assuming one mole of the compound.

**CO-CO **_{2}

_{2}

This example explains in a simple way how the law of multiple proportions works; in CO (carbon monoxide), there are 1,333 grams of oxygen for every gram of carbon. In carbon dioxide (CO₂), there are 2,666 grams of oxygen for every gram of carbon. Therefore, the ratio of oxygen in both compounds is 1: 2, a small integer.

**H **_{2} O-H _{2} O _{2}

_{2}O-H

_{2}O

_{2}

The law of multiple proportions applies to the pair of compounds H _{2} O-H _{2} O _{2} .

In one mole of H _{2} O, 2 grams of hydrogen combine with 16 grams of oxygen. Meanwhile, in one mole of H _{2} O _{2} , 2 grams of hydrogen combine with 32 grams of oxygen. To check if this law is fulfilled, we must set the same mass for one of the elements in both compounds. This time it’s hydrogen: 2 grams.

The mass ratios for O in H _{2} O and H _{2} O _{2} are:

- H
_{2}O: 16 grams O / 2 grams H - H
_{2}O_{2}: 32 grams O / 2 grams H

The mass ratio of O will be 16:32. However, we can simplify it by dividing by 16, remaining 1: 2. Again, the final relation is made up of small whole numbers.

**SO **_{2} -SO _{3}

_{2}-SO

_{3}

In one mole of SO _{2} , 32 grams of sulfur combines with 32 grams of oxygen. Meanwhile, in one mole of SO _{3} , 32 grams of sulfur combine with 48 grams of oxygen. The mass of sulfur is the same for both compounds, so we can directly compare the oxygen ratios:

- SO
_{2}: 32 grams O - SO
_{3}: 48 grams O

Being the ratio of the mass of oxygen between both compounds equal to 32:48 or 1: 1.5. But weren’t there supposed to be whole numbers? The ratio 1: 1.5 (1 / 1.5) can also be written as 2: 3 (0.6), and again we will have small whole numbers: 2 and 3.

Note that we could also have written the ratio as 48:32 or 1.5: 1, the result being 3: 2. The law does not change, only the interpretation of the relationship: There is 1.5 or 3/2 times more oxygen in SO _{3} than in SO _{2} ; which is the same to say that there is 2/3 or 0.6 times less oxygen in SO _{2} than in SO _{3} .

**NO-NO **_{2} -N _{2} O-N _{2} O _{3} -N _{2} O _{5}

_{2}-N

_{2}O-N

_{2}O

_{3}-N

_{2}O

_{5}

The law can also be applied to a number of compounds. Consider then the nitrogen oxides: NO-NO _{2} -N _{2} O-N _{2} O _{3} -N _{2} O _{5} . To be able to evaluate this law in them we must fix a mass of nitrogen: 28 grams. Why? Because NO and NO _{2} have one less nitrogen atom than other oxides:

- 2 (NO): 28 grams N / 32 grams O
- 2 (NO
_{2}): 28 grams N / 64 grams O - N
_{2}O: 28 grams N / 16 grams O - N
_{2}O_{3}: 28 grams N / 48 grams O - N
_{2}O_{5}: 28 grams N / 80 grams O

So let’s forget about nitrogen and focus on the grams of oxygen:

- 2 (NO): 32 grams OR
- 2 (NO
_{2}): 64 grams OR - N
_{2}O: 16 grams O - N
_{2}O_{3}: 48 grams O - N
_{2}O_{5}: 80 grams O

Being the ratio of the masses of O equal to 32: 64: 16: 48: 80. To simplify it, we divide all its numbers by the smallest, 16, leaving it as 2: 4: 1: 3: 5.

That is, there is: 2 times more oxygen in NO than in N _{2} O, 4 times more oxygen in NO _{2} than in N _{2} O, 3 times more oxygen in N _{2} O _{3} than in N _{2} O , and 5 times more oxygen in N _{2} O _{5} than in N _{2} O. Again, we have small whole numbers, ranging from 1 to 5.

__Limitations__

__Limitations__

The law of multiple proportions is not always true. For example, the molar masses or atomic weights of the elements are not whole figures, but have many decimal places. This completely modifies the calculated relationships, which are no longer simple.

Also, the law is not followed for heavy compounds, such as some hydrocarbons. For example, undecane, C _{11} H _{24} , has 1.0083 or 121/120 times more hydrogen than decane, C _{10} H _{22} , whose ratio is made up of 121 and 120, which are more than ten; they are not small integers.

And finally, the law fails with non-stoichiometric compounds, as with many oxides and sulfides.

**Final Thought**

The law of multiple proportions states that when two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in a ratio of a small whole number.